"Good approximation" for the inverse function of $y = x\log_2 x, \hspace{2mm}x>1 $?

Question

I encountered to solve $x$ from $y$ in the equation $y = x\log_2 x ,\hspace{2mm} x>1$, which is known to have no closed form for its inverse function (https://en.wikipedia.org/wiki/Lambert_W_function). Since I'm only interested in how $x$ scales in terms of $y$, just wondering whether there is some good approximation of this inverse function, especially when $y$ is large.

Answer 1

Since $\log x \approx \log y$, one simple approximation is $x \approx \frac{y}{\log_2 y}$. For a precise version of this, let $f(x) = x \log_2 x$. Then we have $$ \lim_{y \to \infty} \frac{f^{-1}(y)}{y / \log_2 y} = 1. $$ Proof: Observe that $$ f\left(\frac{y}{\log_2 y}\right) = y\left(1 - \frac{\log \log y}{\log y}\right). $$ Since $f$ is monotone increasing, this means that $y / \log_2 y$ is an underestimate of $f^{-1}(y)$. So fix an arbitrary $c < 1$; we just need to show that for sufficiently large $y$, $y / \log_2 y > cf^{-1}(y)$. Again using the fact that $f$ is monotone increasing, it is sufficient to show that $f(y / \log_2 y) > f(c f^{-1}(y))$. We have $$ f(c f^{-1}(y)) = cy + (c \log c)f^{-1}(y). $$ For $y$ sufficiently large, this is no more than $c'y$ for $c < c' < 1$. And for $y$ sufficiently large, $1 - \frac{\log \log y}{\log y} > c'$.

Answer 2

Since you already mentioned Lambert function, you know that the inverse of the function is $$x=\frac{y \log (2)}{W(y \log (2))}$$ and, for large values of the argument $z$, an asymptotics is given by $$W(z)= L_1-L_2+\frac {L_2}{L_1}+\frac {L_2(L_2-2)}{2L_1^2}+\cdots$$ where $L_1=\log(z)$ nd $L_2=\log(L_1)$.

Answer 3

This is discussed in N. G. de Bruijn's classic "Asymptotic Methods in Analysis" in the form $xe^x = t$.

It is shown there on pages 33 and following that, if $L_1 = \log t$ and $L_2 = \log \log t$, then $x =L_1-L_2+\frac{L_2}{L_1} +\frac{\frac12 L_2^2-L_2}{L_1^2} +\frac{\frac13 L_2^3-\frac32 L_2^2+O(L_2)}{L_1^3} $.

More accurate results are also stated.